Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int,int> mp;
int inf =  1e9;

vector<int> daikus( vector<vector<int> >&g , int s ){
    vector<int> res( g.size() , inf );
    vector<bool> used( g.size() , false );
    res[s] = 0;
    used[s] = true;
    queue<mp> q;
    q.push( mp( 0, s ) );
    while( !q.empty() ){
	mp now = q.front();
	q.pop();
	int nn = now.second;
	int cost = now.first;
	for(int i=0;i<g[nn].size();i++){
	    int next = g[nn][i];
	    //if( res[ next ] <= cost + 1 ) continue;
	    if( used[next] )continue;
	    res[next] = cost + 1;
	    used[next] = true;
	    q.push( mp( cost + 1 ,next ) );
	}
    }
    return res;
}


int main(){
    int n,m,s,t;
    cin>>n>>m>>s>>t;
    s--,t--;
    vector< vector<int> > g(n);
    for(int i=0;i<m;i++){
	int u,v;
	cin>>u>>v;
	u--,v--;
	g[u].push_back( v );
	g[v].push_back( u );
    }
    vector<int> sd = daikus( g ,s );
    vector<int> td = daikus( g, t );
    vector<ll> cnts( n+1 , 0 );
    vector<ll> cntt( n+1 , 0 );
    int pass = sd[t];
    for(int i=0;i<n;i++) {
	if( sd[i] != inf )cnts[ sd[i] ]++ ;
	if( td[i] != inf )cntt[ td[i] ]++;
    }
    //cout<<"OK"<<endl;
    ll res = 0;
    for(int i=0;i<pass-1;i++){
	res += cnts[i]*cntt[pass-i-2];
    }
    cout<<res<<endl;

    return 0;
}